PHP一些优先级的问题

直接看代码

<?php
echo ‘1‘.print(2)+3,"\n";

不错，就是这么简单，但是很少有人能正确回答

我们执行一下

[[email protected] test]# php -dvld.active=1  test7.php
Finding entry points
Branch analysis from position: 0
Jump found. Position 1 = -2
filename:       /data/www/test/test7.php
function name:  (null)
number of ops:  6
compiled vars:  none
line     #* E I O op                           fetch          ext  return  operands
-------------------------------------------------------------------------------------
   2     0  E >   ADD                                              ~0      2, 3
         1        PRINT                                            ~1      ~0
         2        CONCAT                                           ~2      ‘1‘, ~1
         3        ECHO                                                     ~2
         4        ECHO                                                     ‘%0A‘
   3     5      > RETURN                                                   1

branch: #  0; line:     2-    3; sop:     0; eop:     5; out1:  -2
path #1: 0,
511

让人很诧异吧

换一种写法

<?php
echo ‘1‘,print(2)+3,"\n";

执行结果

[[email protected] test]# php -dvld.active=1  test7.php
Finding entry points
Branch analysis from position: 0
Jump found. Position 1 = -2
filename:       /data/www/test/test7.php
function name:  (null)
number of ops:  6
compiled vars:  none
line     #* E I O op                           fetch          ext  return  operands
-------------------------------------------------------------------------------------
   2     0  E >   ECHO                                                     ‘1‘
         1        ADD                                              ~0      2, 3
         2        PRINT                                            ~1      ~0
         3        ECHO                                                     ~1
         4        ECHO                                                     ‘%0A‘
   3     5      > RETURN                                                   1

branch: #  0; line:     2-    3; sop:     0; eop:     5; out1:  -2
path #1: 0,
151

再换一种

<?php
echo print(2)+3,"\n";

执行结果

[[email protected] test]# php -dvld.active=1  test7.php
Finding entry points
Branch analysis from position: 0
Jump found. Position 1 = -2
filename:       /data/www/test/test7.php
function name:  (null)
number of ops:  5
compiled vars:  none
line     #* E I O op                           fetch          ext  return  operands
-------------------------------------------------------------------------------------
   2     0  E >   ADD                                              ~0      2, 3
         1        PRINT                                            ~1      ~0
         2        ECHO                                                     ~1
         3        ECHO                                                     ‘%0A‘
   3     4      > RETURN                                                   1

branch: #  0; line:     2-    3; sop:     0; eop:     4; out1:  -2
path #1: 0,
51

未完待续……